A few people have asked about the limit comparison test during office hours. It’s stated on page 564 of your text, but we can reword it in a way that might make it easier to understand.
First, imagine that instead of assuming that , we assumed that
for all values of n, not just in the limit.
If , we could write
and
. Either way,
is a multiple of
and vice versa. Multiplying a series by a constant (non-zero, non-infinite) doesn’t change whether it converges or diverges. Therefore, either both series
and
converge or they both diverge….both series have to behave the same way.
If is very very large, then
, that is,
is a tiny multiple of
. Therefore, if you know that
converges, then
must also converge.
If is very very small, then
, that is,
is a tiny multiple of
. Therefore, if you know that
converges, then
must also converge.
None of what I’ve said so far should sound too crazy. What’s interesting about the limit comparison test is that you don’t have to assume that for all values of
, you just have to assume that
is close to
only for large values of
. If
, what this says is that
is eventually almost a multiple of
and vice versa. Therefore, either both series
and
converge or they both diverge…both series have to behave the same way.
Additionally, if you let actually approach
, then
is eventually almost a tiny multiple of
. Therefore, if you know that
converges, then
must also converge.
Finally, if you let actually approach
, then
is eventually almost a tiny multiple of
. Therefore, if you know that
converges, then
must also converge.
tl;dr: If the individual terms of one series are eventually almost equal to a multiple of the individual terms of another series, then both series have to behave the same way (both converge or both diverge).